Free-body diagrams. a. Find the acceleration of mass 2. b. Determine the magnitude of the tension in the string.

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1 Free-body diagrams 1. wo blocks of masses m1 = 5.0 kg and m =.0 kg hang on both sides of an incline, connected through an ideal, massless string that goes through an ideal, massless pulley, as shown below. he friction between the blocks and the incline is negligible. he angle between the incline and the horizontal is θ = 30. m m1 θ a. Find the acceleration of mass. b. Determine the magnitude of the tension in the string. c. If mass 1 starts moving at h = 1.5 m from the ground, what is its velocity when it hits the ground?

2 . A 30-mg spider hangs from two threads from the ceiling as shown below a. Find the tension on each thread. b. he ceiling happens to be the ceiling of an elevator that starts moving up with an acceleration of.5 m/s. Find the new tensions on the threads.

3 3. A ball of mass m = 1. kg attached to one end of a string of length l = m and negligible mass moves in horizontal circles at constant speed as shown in the figure below. he angle between the string and the vertical is θ = 10. l θ m a. Determine the tension on the string. b. What is the period of the ball?

4 c. he string will snap if the tension reaches 15 N. What is the maximum speed the ball can have if the string is not to snap? Answers: 1. a) 5.6 m/s b) 1 N c) 4.1 m/s. a) 0.7 mn, 0.4 mn b) 0.34 mn, 0.30 mn 3. a) 1 N b).8 s c) 3.1 m/s

5 Free-body diagrams 1. wo blocks of masses m1 = 5.0 kg and m =.0 kg hang on both sides of an incline, connected through an ideal, massless string that goes through an ideal, massless pulley, as shown below. he friction between the blocks and the incline is negligible. he angle between the incline and the horizontal is θ = 30. m1 m g N m x y m 1 g θ a. Find the acceleration of mass. It looks like the system will move as follows: m1 down and m up. Block 1, taking positives down: m1g m1a Block, up along incline: mg sin ma hat s two equations with two unknowns (a, ) From the first one, m ( g a ), thus 1 m ( g a) m g sin m a 1 m1 msin a g m m 1 (5.0 kg) (.0 kg)sin 30 (9.8 m/s ) 5.6 m/s 7.0 kg b. Determine the magnitude of the tension in the string. m ( g a) 1 (5.0 kg)( )m/s 1 N

6 c. If mass 1 starts moving at h = 1.5 m from the ground, what is its velocity when it hits the ground? his is 1D constant accelerated motion: v v ah with v v ah 5.6 m/s 1.5 m 4.1 m/s

7 . A 30-mg spider hangs from two threads from the ceiling as shown below y x mg a. Find the tension on each thread. x : cos 30 cos y : sin 40 sin 30 mg 0 1 equations with unknowns (1, ) cos 40 1 cos30 cos 40 1sin 40 1 sin 30 mg cos30 mg 1 sin 40 cos 40 tan 30 6 (3010 kg)(9.8 m/s ) 0.7 mn sin 40 cos 40 tan 30 cos 40 (0.7 mn) 0.4 mn cos30 b. he ceiling happens to be the ceiling of an elevator that starts moving up with an acceleration of.5 m/s. Find the new tensions on the threads. he free-body diagram is the same, but now the acceleration in the vertical direction is not zero: x : cos 30 cos y : sin 40 sin 30 mg ma 1

8 he algebra is almost the same: cos 40 1 cos30 cos 40 1sin 40 1 sin 30 m g a cos30 m g a 1 sin 40 cos 40 tan 30 6 (3010 kg)(1.3 m/s ) sin 40 cos 40 tan mn cos 40 (0.34 mn) 0.30 mn cos30

9 3. A ball of mass m = 1. kg attached to one end of a string of length l = m and negligible mass moves in horizontal circles at constant speed as shown in the figure below. he angle between the string and the vertical is θ = 10. l θ m mg a. Determine the tension on the string. Horizontal: v sin m R Vertical: cos mg 0 1. kg9.8 m/s mg 1 N cos cos10 b. What is the period of the ball? Plugging the result from part a in the equation for the horizontal direction: mg v sin m cos R v gr tan he period is the time to cover a distance πr.

10 R R R period v gr tan gtan R is the radius of the circle, so R lsin, and thus lsin lcos m cos15 period.8 s gtan g 9.8 m/s c. he string will snap if the tension reaches 15 N. What is the maximum speed the ball can have if the string is not to snap? Note that if we change the speed, the angle changes too (it is what we intuitively expect and also we obtained 15. sin v gr tan gl. We must be careful not to assume θ = cos For any angle and speed, the equations in the x and y direction are still formally the same, mg cos mg 0 cos mg 1. kg9.8 m/s 1 1 max cos cos 38.4 max 15 N his corresponds to a maximum speed of v max sin max sin 38.4 gl 9.8 m/s m 3.1 m/s cos cos38.4 max

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